Find the product -a^4(7a^5- 5a+ 5)

DerKrebs [107]

Answer:

In arithmetic and algebra, the cube of a number n is its third power, that is, the result of multiplying three instances of n together. The cube is also the number multiplied by its square: n3 = n × n2 = n × n × n.

Step-by-step explanation:

7 0

2 years ago

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I neeed heeellpp gooot 20 minutttteeessss leffttt

max2010maxim [7]

Answer:

its A

Step-by-step explanation:

F is the midpoint of AA' because E.G. bisects AA'.

4 0

2 years ago

the dimensions of the oven tray is 600mm by 500mm. Each cake tin is 25cm in diameter. How many cake tins can be fit on the oven

gulaghasi [49]

That's 60cm by 50cm

so that would be 4 cake tins

8 0

2 years ago

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4(x+5) = 4x +5x it's probably easy but i'm dumb

Irina18 [472]

4(x+5) = 4x +5x

Simplify by distributing and combining like terms
(4)(x)+(4)(5)=4x+5x
4x+20=4x+5x
4x+20=(4x+5x)
4x+20=9x

Subtract 9x from each side
4x+20−9x=9x−9x
5x+20=0

Subtract 20 from each side
−5x+20−20=0−20
−5x=−20

Divide each side by -5
-5x ÷ -5 = -20 ÷ -5
x = 4

P.S this was actually harder than most algebra problems, so don't beat yourself about it :D

6 0

3 years ago

How do i solve that question?

yawa3891 [41]

a) The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ .

b) The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ .

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ .

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable ( $y, t$ ) in each side of the identity:

$6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t$ (1)

$6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C$

Where $C$ is the integration constant.

By table of integrals we find the solution for each integral:

$-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C$

If we know that $x = 0$ and $y = 1$ , then the integration constant is $C = -2$ .

The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ . $\blacksquare$

b) In this case we need to solve a first order ordinary differential equation of the following form:

$\frac{dy}{dx} + p(x) \cdot y = q(x)$ (2)

Where:

$p(x)$ - Integrating factor
$q(x)$ - Particular function

Hence, the ordinary differential equation is equivalent to this form:

$\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x}$ (3)

The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ . $\blacksquare$

The solution for (2) is presented below:

$y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C$ (4)

Where $C$ is the integration constant.

If we know that $p(x) = -\frac{1}{x}$ and $q(x) = x^{2} + \frac{1}{x}$ , then the solution of the ordinary differential equation is:

$y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C$

$y = x\int {x} \, dx + x\int\, dx + C$

$y = \frac{x^{3}}{2}+x^{2}+C$

If we know that $x = 1$ and $y = -1$ , then the particular solution is:

$y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ . $\blacksquare$

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

3 0

2 years ago